To determine the effect of doubling the molality of a dilute solution on the molal depression constant (Kf), we need to understand the concept of colligative properties in chemistry.
Colligative properties are those physical properties of solutions that depend on the concentration of solute particles, rather than the identity of the solute itself. One such colligative property is freezing point depression, which is described by the formula:
\(\Delta T_f = K_f \times m\)
Here:
From the equation, it is clear that \(\Delta T_f\) is directly proportional to the molality m. However, the molal depression constant K_f is a characteristic property of the solvent and remains constant for a given solvent under constant conditions.
Therefore, even if the molality of the solution is doubled, the value of K_f does not change as it is independent of the concentration of the solution. It depends only on the nature of the solvent.
Thus, the molal depression constant K_f will be Unchanged. Hence, the correct answer is Unchanged.
The freezing point depression constant (\( K_f \)) for water is \( 1.86 \, {°C·kg/mol} \). If 0.5 moles of a non-volatile solute is dissolved in 1 kg of water, calculate the freezing point depression.